Question

answer the questions below about the following function:
$f(x) = \log_{5}(x)$

graph of the function $f(x) = \log_{5}(x)$

\textbf{a)} what is the domain of $f(x)$?

domain

\textbf{b)} what is the range of $f(x)$?

range

\textbf{c)} what is the $x$-intercept of $f(x)$? enter your answer as an ordered pair. if no $x$-intercept exists, enter none.

$x$-intercept

\textbf{d)} is $f(x)$ an increasing or decreasing function?

\bigcirc decreasing

\bigcirc increasing

\textbf{e)} what is the equation of the vertical asymptote?

Explanation:

Part (a)

Step 1: Recall domain of logarithmic functions

The general form of a logarithmic function is \( y = \log_b(x) \), where \( b>0, b
eq1 \). For a logarithmic function, the argument \( x \) must be positive because we can only take the logarithm of a positive number. So, the domain of \( f(x)=\log_5(x) \) is all real numbers \( x \) such that \( x > 0 \).

Step 1: Recall range of logarithmic functions

For the logarithmic function \( y=\log_b(x) \), as \( x \) takes on all positive real values, \( y \) (the output) can take on any real number. This is because we can get any real number as a logarithm by choosing an appropriate positive \( x \). So, the range of \( f(x)=\log_5(x) \) is all real numbers.

Step 1: Find \( x \)-intercept

The \( x \)-intercept is the point where \( y = 0 \) (since on the \( x \)-intercept, the \( y \)-coordinate is 0). So we set \( f(x)=0 \), i.e., \( \log_5(x)=0 \).

Step 2: Solve for \( x \)

Recall that \( \log_b(a)=c \) is equivalent to \( b^c=a \). So, for \( \log_5(x) = 0 \), we have \( 5^0=x \). Since \( 5^0 = 1 \), then \( x = 1 \). So the \( x \)-intercept is the ordered pair \( (1, 0) \).

Answer:

The domain of \( f(x) \) is \( (0, \infty) \) (or \( x>0 \))

Part (b)